What is the integral from 3 to 7 of 700xsqrt(3+2x) ?

The integral from 3 to 7 of 700xsqrt(3+2x) is 14280sqrt(17)-7560

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integral from 3 to 7 of 700xsqrt(3+2x)

\int_{3}^{7} 700 x 3 + 2 x d x

1428017 - 7560

Decimal

51317.94833 \dots

Solution steps

\int_{3}^{7} 700 x 3 + 2 x d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 700 \cdot \int_{3}^{7} x 3 + 2 x d x

Apply u-substitution

: \int_{9}^{17} \frac{u ( u - 3 )}{4} d u

= 700 \cdot \int_{9}^{17} \frac{u ( u - 3 )}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 700 \cdot \frac{1}{4} \cdot \int_{9}^{17} u (u - 3) d u

Expand

u (u - 3) : u^{\frac{3}{2}} - 3 u

= 700 \cdot \frac{1}{4} \cdot \int_{9}^{17} u^{\frac{3}{2}} - 3 u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 700 \cdot \frac{1}{4} (\int_{9}^{17} u^{\frac{3}{2}} d u - \int_{9}^{17} 3 u d u)

\int_{9}^{17} u^{\frac{3}{2}} d u = \frac{578 17 - 486}{5}

\int_{9}^{17} 3 u d u = 3417 - 54

= 700 \cdot \frac{1}{4} (\frac{578 17 - 486}{5} - (3417 - 54))

Simplify

700 \cdot \frac{1}{4} (\frac{578 17 - 486}{5} - (3417 - 54)) : 1428017 - 7560

= 1428017 - 7560

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What is the integral from 3 to 7 of 700xsqrt(3+2x) ?
The integral from 3 to 7 of 700xsqrt(3+2x) is 14280sqrt(17)-7560